Theorem elabg 2682

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of [Quine] p. 44. (Contributed by NM, 14-Apr-1995.)

Hypothesis

Ref	Expression
elabg.1	⊢ (x = A → (φ ↔ ψ))

Assertion

Ref	Expression
elabg	⊢ (A ∈ 𝑉 → (A ∈ {x ∣ φ} ↔ ψ))

Distinct variable groups:   ψ,x   x,A

Allowed substitution hints:   φ(x)   𝑉(x)

Proof of Theorem elabg

Step	Hyp	Ref	Expression
1		nfcv 2175	. 2 ⊢ ℲxA
2		nfv 1418	. 2 ⊢ Ⅎxψ
3		elabg.1	. 2 ⊢ (x = A → (φ ↔ ψ))
4	1, 2, 3	elabgf 2679	1 ⊢ (A ∈ 𝑉 → (A ∈ {x ∣ φ} ↔ ψ))

Syntax hints:   → wi 4   ↔ wb 98   = wceq 1242   ∈ wcel 1390  {cab 2023

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bndl 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019

This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553

This theorem is referenced by:  elab2g  2683  intmin3  3633  finds  4266  elxpi  4304  ovelrn  5591  indpi  6326  peano5nni  7698  peano5setOLD  9400